codility test

Cheng Zheng has solved two test questions on codility, and has a great test of intelligence and ability! (I was hit badly. let's take a look.) if you don't talk about it, you have the opportunity to do two tests on codility and use php to solve it. I wrote two programs, and I thought it was okay. but as soon as I came

Do not say, have the opportunity to do on the codility two test questions, with PHP solution. I wrote two programs that felt good at the time, but came out with a 200 score of 50 points. Wanted God to point out the mistakes in my program, and the solution of the great gods. Don't say anything, first on the topic: 1. You is given a non-empty zero-indexed array a consisting of N integers. Each element of the

Exercises on Codility (14) and exercises on codility 14 (1) TieRopes Given n segments of rope-an array of positive integers and a positive integer K, only two adjacent Ropes can be connected at a time. The length of the rope is the same as the length of the previous rope and the position remains unchanged, how many ropes with at least K length can be formed after such a connection? Data range: N [1 .. 10 ^

Practices on Codility (15) and practices on codility (15) (1) NumberSolitaire A game starts with N grids in a row, with the number 0 .. n-1. At first, the pawn piece is in A [0], and each lattice has an integer (which may be positive or negative ). You are in the grid I, you throw the dice and get the points X = [1 .. 6], and then go to the grid I + X. If the grid does not exist, roll the dice again until t

Exercises on Codility (12) and exercises on codility 12 (1) MinMaxDivision Given a non-negative integer array, each integer is [0 .. m], you want to divide it into K segments (cut K-1 knife), the segment can be empty, each element must belong to a segment, each segment must contain 0 or more continuous elements. The sum of the best and maximum segments is required to be as small as possible, and the smalles

Best time to buy and buy stock Say you have an array for which the ith element is the price of a given stock on day I. IfYou were only permitted to complete at most one transaction(Ie, buy one and every one share of the stock), design an algorithm to find the maximum profit. Ideas Best time to buy and stock seems to be maxprofit in codility. codility puts this question in the chapter of

.. 100,000]; Each element of within the range [−10,000. ]. Complexity: Expected worst-case time complexity is O (N); Expected worst-case space complexity is O (N), beyond input storage (not counting the storage required for input arguments) . Elements of input arrays can be modified.Copyright 2009–2015 by Codility Limited. All rights Reserved. Unauthorized copying, publication or disclosure prohibited.S

above.Assume that: N is an integer within the range [1.. 100,000]; Each element of within the range [−10,000. ]. Complexity: Expected worst-case time complexity is O (N*log (N)); Expected worst-case space complexity is O (N), beyond input storage (not counting the storage required for input arguments) . Elements of input arrays can be modified.Copyright 2009–2015 by Codility Limited. All

set them on peaks 1, 5 and 10; Four flags, you can set only three flags, on peaks 1, 5 and 10. You can therefore set a maximum of three.Write a function: int solution (int a[], int N); That is, given a non-empty zero-indexed array A of N integers, returns the maximum number of flags that can is set on the pea KS of the array.For example, the following array A: A[0] = 1 a[1] = 5 a[2] = 3 a[3] = 4 a[4] = 3 a[5] = 4 a[6] = 1 a[7] = 2 a[8] = 3

https://codility.com/programmers/challenges/fluorum2014http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1273http://blog.csdn.net/caopengcs/article/details/36872627http://www.quora.com/ how-do-i-determine-the-order-of-visiting-all-leaves-of-a-rooted-tree-so-that-in-each-step-i-visit-a-leaf-whose-path-from- root-contains-the-most-unvisited-nodes#Thinking as Cao Bo said, first DFS records the distance from the root, the direction of the precursor. Then sort by distance, then sort by th

integer within the range [0... 1,000,000,000]. Complexity: Expected worst-case time complexity is O (n * log (N ))); Expected worst-case space complexity is O (n), Beyond Input Storage (not counting the storage required for input arguments ). Elements of input Arrays can be modified.Ideas The general idea is that we want to divide the sequence into C slices. Each slice must have at least one peak, and the peak is a number larger than the left and right sides (the first and last par

Connection to the Brackets of codility: BracketsQuestion analysis It is actually a balanced symbol.How can this problem be solved? You can use the stack to get it done. for details, read the book.Talk is cheap, show me the code ~ Code: Function solution ($ S) {// write your code in PHP5.5 $ open_symb = array ('{','[','('); $ close_symb = array ('}', ']', '); $ check = array (); for ($ I = 0; $ I

I haven't written the codility question for a long time, so I have no time, and I am not very good at analyzing some questions. This question is interesting, and I have not provided a very strict proof. Given a tree (undirected acyclic graph), start from a node and select as many nodes as possible from the start point to the destination node path, until all nodes are traversed. If there are as many nodes in the path from the start point to the two d

Question:https://codility.com/programmers/lessons/4We need, parts to prove, our solution.On one hand,there is no false triangular. Given The array has been sorted, if a[i]+a[i+1]>a[i+2], we can prove the existence of the triangle. For array A are sorted, we can easily confirm that a[i+2] + a[i] > a[i+1] and a[i+1]+a[i+2] >a[i]. So we just need to check this condition.On the other hand,There is no underreporting triangular.If The inequality can hold for three out-of-order elements, to say, A[inde

convert if(Avg min) {min=avg; Res=Left ; } if(A[left] > Avg left //When move left pointer can decrease the AVG then move it.left++; Sumleft+ = A[left-1]; Continue //must have this, or may appear the right grow to Len in next step and left have no chance to update. } if(Right //When move right pointer can decrease the AVG then move it.|| left = = right-1) {//Make sure threr are not "piece/0"right++; if(Right len) sum+=A[right]; } Else{ Left++;//if not

classSolution { Public intSolution (int[] A) {//Write your code in Java SE 8 if(a.length){ return0; }Else{ intSorted[] =Bubblesort (A); for(inti=0;i){ if(i==sorted.length-1){ return0; }Else{ intdifference = sorted[i+1]-Sorted[i]; if(difference==2){ returnSorted[i]+1; } } } } return0; } Public int[] Bubblesort (int[] number) { inttemp = 0

there are lotus leaves in these locations, 0 means no, and 1 indicates there are. Frogs from the shore to jump to the lotus leaf, through 0 or more lotus leaf jump to the other side, and only to the opposite side of the direction of jumping-can't jump back. And the frog each jump distance must be a Fibonacci number, ask the frog at least a few steps to jump to the opposite shore? No solution returns-1.Data range: N [0..30000]Complexity required: Time O (Nlogn), Space O (N).Analysis: First we ca

(1) NumbersolitaireA game starts with a row of n squares, with a lattice number 0. N-1, at first, the pieces in a[0], each lattice has an integer (possibly positive, possibly negative). You are in lattice I, you throw the dice, get the points x = [1..6], and then go to the lattice numbered i + x, if this lattice does not exist, throw a dice again until i + x lattice exists. When you go to the N-1th grid, the game is over. The number of integers you pass through the lattice is your score, and the

duplicate numbers. If the result exceeds 10^9, the 10^9 is returnedData range: M [0..10^5], array length n [1..10^5]Requires a complexity of time O (N), Space O (M).Analysis: This problem is very much like a substring containing only one character, but the difference is that we are asking for the number. We use O (M) space to maintain a sliding window where the values are not duplicated. For each I, we find the J satisfies the window [I.. J] elements are not duplicated, then the number of non-r

Codility adds exercise lesson 2. There are three questions: (1) Perm-Check the given integer array has N numbers. Ask if it is an arrangement of 1-N, that is, whether each number is 1-N and only appears once. Output 1 and 0 indicate whether or not, input range N [1 .. 10 ^ 5], an integer in the array [1 .. 10 ^ 5], requires the complexity of the time space is O (N ). Analysis: the space complexity O (N) algorithm is very simple. We can create a bool a